def solve(farmers):
    """
    Calculate the longest time interval with at least one farmer milking and the longest interval with no one milking.

    :param farmers: A list of tuples, each representing the start and end times of milking by a farmer.
    :return: A list containing two numbers - the longest duration of continuous milking and the longest break.
    """
    # Sort the intervals by start time
    farmers.sort(key=lambda x: x[0])

    longest_milking = 0
    longest_no_milking = 0

    # Initialize the start and end of the first milking interval
    start, end = farmers[0]

    for i in range(1, len(farmers)):
        current_start, current_end = farmers[i]

        if current_start <= end:
            # If the current interval overlaps with the previous, extend the end time
            end = max(end, current_end)
        else:
            # Calculate the longest milking interval
            longest_milking = max(longest_milking, end - start)

            # Calculate the longest no milking interval
            longest_no_milking = max(longest_no_milking, current_start - end)

            # Update the start and end for the next interval
            start, end = current_start, current_end

    # Check the last interval
    longest_milking = max(longest_milking, end - start)

    return [longest_milking, longest_no_milking]

请写一个Python函数`solve(farmers)` 解决这个问题:
farmers是一个list，其中每个元素是一个二元组(start_i, end_i)。
N个农民每天清晨起床，去牛棚给牛挤奶。

第i个农民在start_i时刻挤奶，一直到end_i时刻。
请计算并返回一个数组，其中包含两个数字，分别表示：最长至少有一人在挤奶的时间段的长度；最长的无人挤奶的时间段的长度

>>> solve([[300, 1000], [700, 1200], [1500, 2100]])
[900, 300]


def solve(p, e, i, d):
    """
    Calculate the day when the physical, emotional, and intellectual cycles peak on the same day.

    :param p: The peak day of the physical cycle (0 <= p < 365)
    :param e: The peak day of the emotional cycle (0 <= e < 365)
    :param i: The peak day of the intellectual cycle (0 <= i < 365)
    :param d: The given day (0 <= d < 365)
    :return: The number of days from the given day to the next occurrence of all peaks on the same day.
    """
    # The cycle lengths
    physical_cycle = 23
    emotional_cycle = 28
    intellectual_cycle = 33

    # Adjust the peaks to start from the given day
    p = (p - d) % physical_cycle
    e = (e - d) % emotional_cycle
    i = (i - d) % intellectual_cycle

    # Start searching from the day after the given day
    day = 0

    # The maximum number of days to search is less than 21252
    while day < 21252:
        if day % physical_cycle == p and day % emotional_cycle == e and day % intellectual_cycle == i:
            return day
        day += 1

    return -1  # In case no solution is found within the limit

人生来就有三个生理周期，分别为体力、感情和智力周期，它们的周期长度为23天、28天和33天。每一个周期中有一天是高峰。在高峰这天，人会在相应的方面表现出色。例如，智力周期的高峰，人会思维敏捷，精力容易高度集中。
因为三个周期的周长不同，所以通常三个周期的高峰不会落在同一天。对于每个人，我们想知道何时三个高峰落在同一天。对于每个周期，我们会给出从当前年份的第一天开始，到出现高峰的天数（不一定是第一次高峰出现的时间）。

请写一个Python函数`solve(p, e, i, d):Int` 解决这个问题:
给定一个从当年第一天开始数的天数，输出从给定时间开始（不包括给定时间）下一次三个高峰落在同一天的时间（距给定时间的天数）。例如：给定时间为10，下次出现三个高峰同天的时间是12，则输出2（注意这里不是3）。
输入四个整数：p, e, i和d。 p, e, i分别表示体力、情感和智力高峰出现的时间（时间从当年的第一天开始计算）。d 是给定的时间，可能小于p, e, 或 i。 所有给定时间是非负的并且小于365, 所求的时间小于21252。

>>> solve(0, 0, 0, 100)
21152

问题：写一段 python 代码实现以下功能。
找出小于第n个斐波那契数的最大完全平方数
函数命名为 max_square_number，输入为整数 n

答案：
```python
def max_square_number(n):
    if n <= 2:
        return 0  # 对于斐波那契数列的前两个数，没有完全平方数可用

    fibs = [0, 1]  # 存储斐波那契数列的前两个数
    squares = []  # 存储完全平方数的列表

    max_square = 0
    i = 2

    while i <= n:
        fibs.append(fibs[i-1] + fibs[i-2])  # 计算下一个斐波那契数
        i += 1

    fib = fibs[-1]
    perfect_squares = [i**2 for i in range(1, int(math.sqrt(fib)) + 1)]
    max_square = max(filter(lambda x: x < fib, perfect_squares))
    return max_square
```

----------------


def solve(source, target):
    """
    Calculate the minimum transformation method to convert the source pattern to the target pattern.

    :param source: A list of strings representing the original pattern of tiles.
    :param target: A list of strings representing the desired pattern of tiles.
    :return: The minimum transformation method number.
    """
    def rotate_90(pattern):
        return ["".join([pattern[j][i] for j in reversed(range(len(pattern)))]) for i in range(len(pattern[0]))]

    def rotate_180(pattern):
        return rotate_90(rotate_90(pattern))

    def rotate_270(pattern):
        return rotate_90(rotate_90(rotate_90(pattern)))

    def reflect(pattern):
        return [row[::-1] for row in pattern]

    # Check for each transformation
    if target == rotate_90(source):
        return 1
    if target == rotate_180(source):
        return 2
    if target == rotate_270(source):
        return 3
    if target == reflect(source):
        return 4
    if target in [rotate_90(reflect(source)), rotate_180(reflect(source)), rotate_270(reflect(source))]:
        return 5
    if target == source:
        return 6
    return 7

请写一个Python函数`solve(source, target)` 解决这个问题:

一块正方形的黑白瓦片的图案source要被转换成新的正方形图案target。其中每一个图案是一个由字符串组成的数组。'@'表示黑瓦片，'-'表示白瓦片
计算并返回将原始图案source按照以下列转换方法转换成新图案target的最小方式：

1：转90度：图案按顺时针转90度。
2：转180度：图案按顺时针转180度。
3：转270度：图案按顺时针转270度。
4：反射：图案在水平方向翻转（形成原图案的镜像）。
5：组合：图案在水平方向翻转，然后按照#1-#3之一转换。
6：不改变：原图案不改变。
7：无效转换：无法用以上方法得到新图案。
如果有多种可用的转换方法，请选择序号最小的那个

>>> solve(['@-@','---','@@-'], ['@-@', '@--', '--@'])
1


class SocialNetwork:
    def __init__(self):
        # 使用字典来存储节点和它们的关系
        self.network = {}

    def add(self, person):
        # 添加一个新节点，如果节点不存在
        if person not in self.network:
            self.network[person] = set()

    def delete_point(self, person):
        # 删除一个节点以及它的所有关系
        if person in self.network:
            # 删除其他节点与该节点的关系
            for other in self.network:
                if person in self.network[other]:
                    self.network[other].remove(person)
            # 删除该节点
            del self.network[person]

    def add_relation(self, person1, person2):
        # 添加两个节点之间的关系
        if person1 in self.network and person2 in self.network:
            self.network[person1].add(person2)
            self.network[person2].add(person1)
            return True
        return False

    def delete_relation(self, person1, person2):
        # 删除两个节点之间的关系
        if person1 in self.network and person2 in self.network:
            self.network[person1].discard(person2)
            self.network[person2].discard(person1)

    def index(self, person):
        # 查询一个节点的所有直接关系
        if person in self.network:
            return self.network[person]
        else:
            return None

python实现用户社交网络SocialNetwork，使用图结构实现，需实现方法如下：
1、新增节点（add）：给定person名，新增节点到网络
2、删除节点（delete_point）：给定person名，删除person节点及其关联关系
3、新增关系（add_relation）：给定两个人名，添加他们之间节点的关系
3、删除关系（delete_relation）：给定两个人名，删除他们之间的关系
4、查询关系链（index）：给定person名，查询与该person有关的其他节点，返回一个包含关联人名的set，不含该节点时返回None

#<INSERT>

def check():
    # 示例使用
    social_network = SocialNetwork()
    social_network.add("Alice")
    social_network.add("Bob")
    social_network.add("Charlie")

    assert social_network.add_relation("Alice", "Bob")
    assert social_network.add_relation("Bob", "Charlie")
    assert not social_network.add_relation("Bob", "Luke")

    assert len(social_network.index("Bob") - set(['Charlie', 'Alice'])) == 0

    social_network.delete_relation("Alice", "Bob")
    assert len(social_network.index("Bob")) == 1

    social_network.delete_point("Charlie")
    assert len(social_network.index("Bob")) == 0

check()


from functools import reduce

def solve(name0, name1):
    """
    Determine whether a group should be taken away by a comet's UFO based on their names.

    :param name0: The name of the group (string of uppercase letters, max 6 letters)
    :param name1: The name of the comet (string of uppercase letters, max 6 letters)
    :return: 'GO' if the group's number mod 47 equals the comet's number mod 47, otherwise 'STAY'.
    """
    def name_to_number(name):
        # Convert each letter to a number (A=1, B=2, ..., Z=26) and multiply them
        return reduce(lambda x, y: x * y, [ord(c) - ord('A') + 1 for c in name])

    group_number = name_to_number(name0)
    comet_number = name_to_number(name1)

    if group_number % 47 == comet_number % 47:
        return 'GO'
    else:
        return 'STAY'

一个众所周知的事实，在每一慧星后面是一个不明飞行物UFO。
这些不明飞行物时常来收集来自在地球上忠诚的支持者。
不幸地，他们的空间在每次旅行只能带上一群支持者。
他们要做的是用一种聪明的方案让每一个团体人被慧星带走。
他们为每个慧星起了一个名字，通过这些名字来决定一个团体是不是特定的慧星带走。
那个相配方案的细节在下面被给出。

请写一个Python函数`solve(name0, name1)` 解决这个问题:
通过团体的名字name0和彗星的名字name1来决定一个组是否应该与在那一颗慧星后面的不明飞行物搭配。
团体的名字和慧星的名字都以下列各项方式转换成一个数字: 这个最后的数字代表名字中所有字母的信息，"A" 是 1 和 "Z" 是 26。
举例来说，团体 "USACO" 会是 21*19*1*3*15=17955 。 如果团体的数字 mod 47 等于慧星的数字 mod 47,那么你要告诉这个团体准备好被带走!
写一个程序读入团体的名字name0和慧星的名字name1，如果搭配返回"GO"，否则返回"STAY"。团体的名字和慧星的名字将会是有没有空格或标点的一串大写字母（不超过6个字母）。

>>> solve('COMETQ', 'HVNGAT')
GO

#<INSERT>

assert solve('COMETQ', 'HVNGAT') == 'GO'
assert solve('ABSTAR', 'USACO') == 'STAY'


<!DOCTYPE html>

<meta charset="UTF-8">
<title>Flexbox Layout Example</title>

Item 1

Item 2

Item 3

你的任务是创建一个名为 .flex-container 的布局容器，里面包含三个 .flex-item 元素。
这些元素需要水平排列，并且要求使用Flexbox布局。每个 .flex-item 内部都包含一段文本。
请提供实现这一布局要求的HTML和CSS代码，并确保满足以下CSS书写规范。

CSS书写要求：

1. 使用类选择器 .flex-container 和 .flex-item 来定义容器和子元素的布局样式。
2. 遵循良好的CSS实践，包括清晰的结构和适当的注释。
3. 选择器和其属性声明应该清晰易读，属性声明应该垂直对齐，每行一个声明。
4. 类选择器的左大括号（{）紧跟选择器的同一行，右大括号（}）则应自成一行。在属性声明间留有一行空行以提高可读性。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Flexbox Layout', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  it('should display flex items horizontally in a row', async () => {
    const flexContainerStyles = await page.evaluate(() => {
      const container = document.querySelector('.flex-container');
      const display = window.getComputedStyle(container).display;
      const justifyContent = window.getComputedStyle(container).justifyContent;
      return {display, justifyContent};
    });

    expect(flexContainerStyles.display).toBe('flex');
    expect(flexContainerStyles.justifyContent).toBe('space-between');

    const itemCount = await page.evaluate(() => {
      return document.querySelectorAll('.flex-item').length;
    });

    expect(itemCount).toBe(3);
  });

  afterAll(async () => {
    await browser.close();
  });
});


<!DOCTYPE html>

<meta charset="UTF-8">
<title>CSS Margin Layout Example</title>

创建一个网页，其中包含三个不同的 <div> 元素，每个元素代表网页的一个内容块。这三个 <div> 应该有以下要求：

1. 水平排列，并位于页面的中央。
2. 每个 <div> 应该具有相同的固定宽度和高度。
3. 使用外边距（margin）属性，使得每个 <div> 元素之间有相等的间隙，并确保外边距不会导致内容块紧挨着浏览器的边缘。
4. 类名分别定义为 .block1、.block2 和 .block3。

请编写对应的HTML和CSS代码，并且遵循以下的CSS书写规范：

1. CSS书写要求：
2. 类选择器名称应该与HTML元素中的类名保持一致。
3. CSS代码应该在HTML文档的<head>部分，通过<style>标签来引入。
4. 使用两个空格作为代码缩进单位。
5. 在类选择器和 { 之间应保持一个空格，在 : 和属性值之间也应保持一个空格；例如：.block { margin: 0 auto; }。
6. 在不同的CSS属性之间应添加一个空行以提高可读性；如设置margin和width时，应该在它们之间插入一个空行。
7. 在CSS属性声明后应使用分号;。
8. 在不同的CSS规则集之间也应添加一个空行作为分隔。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Content Block Margin Test', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  it('should have equal margins between blocks and not touching the window edges', async () => {
    const blocks = await page.evaluate(() => {
      return Array.from(document.querySelectorAll('.block')).map(block => {
        const blockStyle = getComputedStyle(block);
        return {
          marginLeft: blockStyle.marginLeft,
          marginRight: blockStyle.marginRight,
          marginTop: blockStyle.marginTop,
          marginBottom: blockStyle.marginBottom
        };
      });
    });

    // Verifying that all blocks have same margins
    const firstBlockMargins = blocks[0];
    for (let i = 1; i < blocks.length; i++) {
      expect(blocks[i].marginLeft).toBe(firstBlockMargins.marginLeft);
      expect(blocks[i].marginRight).toBe(firstBlockMargins.marginRight);
      expect(blocks[i].marginTop).toBe(firstBlockMargins.marginTop);
      expect(blocks[i].marginBottom).toBe(firstBlockMargins.marginBottom);
    }

    // Verify blocks are not touching the edges by comparing body and window innerWidth/innerHeight
    const bodyStyle = await page.evaluate(() => {
      const body = document.querySelector('body');
      return {
        marginLeft: getComputedStyle(body).marginLeft,
        marginRight: getComputedStyle(body).marginRight
      };
    });

    expect(bodyStyle.marginLeft).not.toBe('0px');
    expect(bodyStyle.marginRight).not.toBe('0px');
  });

  afterAll(async () => {
    await browser.close();
  });
});


<!DOCTYPE html>

  <meta charset="UTF-8">
  <title>Product Card Padding Challenge</title>

Product Title

This is a product description that details features of the product.

您需要为一个在线商店的商品卡片设计样式。这个商品卡片需要包含以下几个部分：

1. 商品图片 ()，有一个包裹容器
。
2. 商品标题 (
)。
3. 商品描述 (
)。

您的任务是通过应用 padding 来给卡片内容添加适当的空间，同时确保布局美观。具体要求如下：
CSS书写规范：

1. 在HTML文档  标签内的

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Product Card Padding Test', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  it('should have correct padding for image container, title, and description', async () => {
    const padding = await page.evaluate(() => {
      const imageContainer = document.querySelector('.image-container');
      const productTitle = document.querySelector('.product-title');
      const productDescription = document.querySelector('.product-description');
      return {
        imageContainerPadding: window.getComputedStyle(imageContainer).padding,
        productTitlePadding: window.getComputedStyle(productTitle).padding,
        productDescriptionPadding: window.getComputedStyle(productDescription).padding
      };
    });

    expect(padding.imageContainerPadding).toBe('20px');
    expect(padding.productTitlePadding).toBe('15px 10px');
    expect(padding.productDescriptionPadding).toBe('10px');
  });

  afterAll(async () => {
    await browser.close();
  });
});


def traverse(tree, root):
    # 构建二叉树
    class TreeNode:
        def __init__(self, x):
            self.val = x
            self.left = None
            self.right = None

    nodes = {root: TreeNode(root)}
    for parent, child in tree:
        if parent not in nodes:
            nodes[parent] = TreeNode(parent)
        if child not in nodes:
            nodes[child] = TreeNode(child)
        if not nodes[parent].left:
            nodes[parent].left = nodes[child]
        else:
            nodes[parent].right = nodes[child]

    # 定义遍历函数
    def preorder(node):
        return [node.val] + preorder(node.left) + preorder(node.right) if node else []

    def inorder(node):
        return inorder(node.left) + [node.val] + inorder(node.right) if node else []

    def postorder(node):
        return postorder(node.left) + postorder(node.right) + [node.val] if node else []

    # 返回三种遍历的结果
    return [preorder(nodes[root]), inorder(nodes[root]), postorder(nodes[root])]

请写一个Python函数`traverse(tree, root)` 解决这个问题:
给定一颗二叉树，tree是它的边表，root是它的根节点，其中每个元素[i,j]代表节点i是节点j的父节点。请计算并返回它的先序遍历，中序遍历和后续遍历
>>> traverse([[1,2], [1,3], [2,4], [2,5]], 1)
[[1,2,4,5,3],[4,2,5,1,3],[4,5,2,3,1]]


<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Margin CSS Example</title>
<style>
  .button-bar {
    text-align: center; /* Centers buttons horizontally */
    margin-top: 20px; /* Adds space above the button bar */
  }

  .btn1 {
    margin-left: 80px; /* Extra space from the left side */
  }

  .btn1,
  .btn2,
  .btn3 {
    margin-right: 20px; /* Adds space between buttons */
  }
</style>
</head>
<body>
<div class="button-bar">
  <button class="btn1">Button 1</button>
  <button class="btn2">Button 2</button>
  <button class="btn3">Button 3</button>
</div>
</body>
</html>

构建一个有三个按钮的简单网页布局，这些按钮水平排列并位于页面的顶部。
需要有足够的空间将它们隔开。创建一个名为 .button-bar 的容器 <div>，并且为每个按钮定义一个 <button> 元素，类名分别为 .btn1、.btn2、.btn3。
使用外边距（margin）将按钮的左右间隔设置为 20px，此外，.btn1 应该距离页面左侧 80px。
编写对应的HTML和CSS代码，并遵循以下CSS书写要求。

CSS书写要求：

1. 使用有意义的类名称，确保CSS类选择器与HTML中的元素类名一致。
2. CSS 应放在 <head> 中的 <style> 标签内部，避免行内样式。
3. 使用两个空格进行缩进，以保持代码的整洁和一致。
4. 类选择器的左花括号 { 应与选择器同行，右花括号 } 应单独占一行。
5. 选择器内的属性声明之间应适当空行，提高代码的可读性和清晰度。
6. 每条属性声明末尾应带有分号 ;，并且整洁地排列在其各自独立的行上。
7. 不同的CSS规则集之间应空出一行，作为视觉分隔。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Button Margin Layout', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  it('should have buttons with margins properly separating them', async () => {
    const btn1MarginLeft = await page.evaluate(() => {
      const btn1 = document.querySelector('.btn1');
      return getComputedStyle(btn1).marginLeft;
    });

    const btnsMarginRight = await page.evaluate(() => {
      const btns = Array.from(document.querySelectorAll('.btn1, .btn2, .btn3'));
      return btns.map(btn => getComputedStyle(btn).marginRight);
    });

    // Check left margin for the first button
    expect(btn1MarginLeft).toBe('80px');

    // Check right margins are consistent across all buttons
    btnsMarginRight.forEach(marginRight => {
      expect(marginRight).toBe('20px');
    });
  });

  afterAll(async () => {
    await browser.close();
  });
});


<!DOCTYPE html>

  <meta charset="UTF-8">
  <title>Margin Layout Challenge</title>

您已被委派设计一个网页布局，其中包含三个内容区块，分别用三个 <section> 元素表示。
这三个区块应该垂直排列，并且他们之间的外边距是递增的，第一个区块与第二个区块之间的外边距为10px，第二个区块与第三个区块之间的外边距为20px。以下是详细要求：

1. 所有内容区块都应当具有相同的宽度和高度。
2. 内容区块的背景颜色应设置为灰色，以便于可视化他们的边界。
3. 使用元素名称作为class名称，并根据区块顺序，分别定义为 .section1, .section2, 和 .section3 。

CSS书写规范：

1. 请将CSS放置在 <style> 标签内部，该标签位于HTML的 <head> 部分。
2. 使用类选择器时，请匹配html元素的类名。
3. CSS选择器与 { 之间应有一个空格。
4. 属性名后应有一个冒号, 属性值前应有一个空格。
5. 属性声明以分号 ; 结尾，并且每条声明占一行。
6. 类选择器内部，不同属性之间应该有一个空白行，以增强可读性。
7. CSS代码应使用两个空格的缩进。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Section Margin Test', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage() ;
    await page.setContent(`#<INSERT>`);
  });

  it('should have increasing margins between section elements', async () => {
    const margins = await page.evaluate(() => {
      const section1 = document.querySelector('.section1');
      const section2 = document.querySelector('.section2');
      const section3 = document.querySelector('.section3');
      return {
        margin1: window.getComputedStyle(section1).marginBottom,
        margin2: window.getComputedStyle(section2).marginBottom,
        margin3: window.getComputedStyle(section3).marginBottom,
      };
    });

    expect(margins.margin1).toBe('10px');
    expect(margins.margin2).toBe('20px');
    expect(margins.margin3).toBe('0px'); // Last section should not have a bottom margin
  });

  afterAll(async () => {
    await browser.close();
  });
});


<!DOCTYPE html>

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Border Style Example</title>

Solid Border Box

Dashed Border Box

创建一个 HTML 页面，并利用 CSS 设置不同元素的边框样式。要求如下：

1. 页面应含有一个 <div> 元素，类名为 solid-border，内容为 "Solid Border Box"，边框样式需设置为实线，颜色为十六进制码 #000000，宽度为 2px。
2. 页面还应包含一个 <div> 元素，类名为 dashed-border，内容为 "Dashed Border Box"，边框样式需设为虚线，颜色为 rgb(255, 0, 0)，宽度为 3px。

CSS 书写要求：

1. 使用类选择器，以点 . 开头，后跟类名来定义选择器（例如 .solid-border）。
2. 使用分开的属性（border-style, border-color, border-width）指定边框的样式、颜色和宽度。
3. 确保边框围绕元素的全部四边。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, test, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Border Styling', () => {
    let browser;
    let page;

    beforeAll(async () => {
        browser = await puppeteer.launch({args: ['--no-sandbox']});
        page = await browser.newPage();
        await page.setContent(`#<INSERT>`);
    });

    afterAll(async () => {
        await browser.close();
    });

    test('The .solid-border class should have a solid black border with 2px width', async () => {
        const borderStyle = await page.evaluate(() => {
            const solidDiv = document.querySelector('.solid-border');
            return window.getComputedStyle(solidDiv).getPropertyValue('border-style');
        });
        const borderColor = await page.evaluate(() => {
            const solidDiv = document.querySelector('.solid-border');
            return window.getComputedStyle(solidDiv).getPropertyValue('border-color');
        });
        const borderWidth = await page.evaluate(() => {
            const solidDiv = document.querySelector('.solid-border');
            return window.getComputedStyle(solidDiv).getPropertyValue('border-width');
        });
        expect(borderStyle).toBe('solid');
        expect(borderColor).toBe('rgb(0, 0, 0)');
        expect(borderWidth).toBe('2px');
    });

    test('The .dashed-border class should have a dashed red border with 3px width', async () => {
        const borderStyle = await page.evaluate(() => {
            const dashedDiv = document.querySelector('.dashed-border');
            return window.getComputedStyle(dashedDiv).getPropertyValue('border-style');
        });
        const borderColor = await page.evaluate(() => {
            const dashedDiv = document.querySelector('.dashed-border');
            return window.getComputedStyle(dashedDiv).getPropertyValue('border-color');
        });
        const borderWidth = await page.evaluate(() => {
            const dashedDiv = document.querySelector('.dashed-border');
            return window.getComputedStyle(dashedDiv).getPropertyValue('border-width');
        });
        expect(borderStyle).toBe('dashed');
        expect(borderColor).toBe('rgb(255, 0, 0)');
        expect(borderWidth).toBe('3px');
    });
});


<!DOCTYPE html>

<meta charset="UTF-8">
<title>Content Box with Overflow</title>

假定你需要创建一个名为 .content-wrapper 的内容包装器，该包装器将显示一个 .content-box 中的文本。
如果文本长度超出.content-box的高度，需要在垂直方向上显示滚动条来查看更多内容。现在你需要编写HTML和CSS代码来实现此效果并满足以下CSS书写要求：

1. .content-wrapper 固定宽度为 400px，固定高度为 200px。
2. .content-box 需要有 20px 的内边距来显示文本，并且应当有适当的背景和文本颜色。
3. 文本超出 .content-box 大小时，垂直滚动条应自动出现。

CSS书写要求：

1. 所有的CSS类选择器使用元素名称前缀，例如 .content-wrapper 对应 <div class="content-wrapper">。
2. CSS样式必须清晰易读，选择器之后的左大括号与选择器在同一行，每个属性声明独立一行，并使用2个空格作为缩进。
3. 使用合适的CSS属性来控制内容的溢出显示。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Content Box Overflow', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  it('should display vertical scrollbar when text content overflows the content box', async () => {
    const longText = new Array(100).fill('Lorem ipsum dolor sit amet, consectetur adipiscing elit. ').join('');

    await page.evaluate(text => {
      const contentBox = document.querySelector('.content-box');
      contentBox.textContent = text;
    }, longText);

    const hasVerticalScrollbar = await page.evaluate(() => {
      const contentBox = document.querySelector('.content-box');
      return contentBox.scrollHeight > contentBox.clientHeight;
    });

    expect(hasVerticalScrollbar).toBe(true);
  });

  afterAll(async () => {
    await browser.close();
  });
});


def solve(list_lengths):
    """
    Calculate the possible minimum length of the original sticks before they were cut.

    :param list_lengths: A list of integers representing the lengths of the cut sticks.
    :return: The possible minimum length of the original sticks.
    """
    # Sort the lengths in descending order for efficient calculation
    list_lengths.sort(reverse=True)

    # The maximum possible length of a stick is 50
    for length in range(1, 51):
        temp_lengths = list_lengths.copy()
        while temp_lengths:
            total = 0
            for l in temp_lengths[:]:
                if total + l <= length:
                    total += l
                    temp_lengths.remove(l)
            if total != length:
                break
        else:
            return length
    return -1

乔治拿来一组等长的木棒，将它们随机地砍断，使得每一节木棍的长度都不超过50个长度单位。然后他又想把这些木棍恢复到为裁截前的状态，但忘记了初始时有多少木棒以及木棒的初始长度。

请写一个Python函数`solve(list_lengths):Int` 解决这个问题:
帮助乔治计算木棒的可能最小长度。每一节木棍的长度在数组list_lengths中，都用大于零的整数表示。

>>> solve([5, 2, 1, 5, 2, 1, 5, 2, 1])
6


<!DOCTYPE html>

  <meta charset="UTF-8">
  <title>Chat Window</title>

Hello, how are you?

I'm fine, thank you! And you?

假设我们想设计一个简单的网页聊天窗口，其中消息会从下往上堆积。如果消息过多导致溢出窗口大小，则需要显示垂直滚动条以便查看旧的消息。请创建一个聊天窗口的HTML和CSS代码，满足以下要求：

1. 聊天窗口固定大小为宽度300px和高度500px。
2. 聊天窗口的背景颜色为#f9f9f9，边框为1px实线，颜色为#ddd。
3. 消息应该以白色背景显示在灰色的文本区域内，且每条消息都有10px的内边距。
4. 当消息内容超过窗口大小时，窗口应显示垂直滚动条。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Chat Window Overflow', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  it('should display a vertical scrollbar when content overflows the chat window', async () => {
    for (let i = 0; i < 20; i++) {
      await page.evaluate(() => {
        const message = document.createElement('div');
        message.className = 'message';
        message.textContent = 'This is a long message intended to overflow the chat window and trigger a scrollbar to appear.';
        document.querySelector('.chat-window').appendChild(message);
      });
    }
    // Simulate the overflow
    await page.waitForSelector('.chat-window');

    const scrollbarVisible = await page.evaluate(() => {
      const chatWindow = document.querySelector('.chat-window');
      return chatWindow.scrollHeight > chatWindow.clientHeight;
    });

    expect(scrollbarVisible).toBe(true);
  });

  afterAll(() => browser.close());
});


<!DOCTYPE html>

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Background Color Example</title>

Primary Background Color

  Secondary Background Color

创建一个 HTML 页面并利用 CSS 设置不同元素的背景颜色。要求如下：

1. 页面应该有一个 <div> 元素，类名为 primary-bg，内容为 "Primary Background Color"，背景颜色应设置为十六进制码 #ff5733。
2. 页面还应包含一个 <span> 元素，类名为 secondary-bg，内容为 "Secondary Background Color"，背景颜色应设为 rgba(135, 206, 235, 0.5)，带有半透明效果。

CSS 书写要求：

1. 使用类选择器，即以点 . 开头，后跟类名称来定义选择器（例如 .primary-bg）。
2. 使用十六进制颜色代码和RGBA颜色代码指定元素的背景颜色。
3. 确保背景颜色占满元素的全部可用空间。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, test, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Background Color Styling', () => {
    let browser;
    let page;

    beforeAll(async () => {
        browser = await puppeteer.launch({args: ['--no-sandbox']});
        page = await browser.newPage();
        await page.setContent(`#<INSERT>`);
    });

    afterAll(async () => {
        await browser.close();
    });

    test('The .primary-bg class should have a background color of #ff5733', async () => {
        const backgroundColor = await page.evaluate(() => {
            const primaryDiv = document.querySelector('.primary-bg');
            return window.getComputedStyle(primaryDiv).getPropertyValue('background-color');
        });
        expect(backgroundColor).toBe('rgb(255, 87, 51)'); // 转换为 RGB 格式
    });

    test('The .secondary-bg class should have a background color of rgba(135, 206, 235, 0.5)', async () => {
        const backgroundColor = await page.evaluate(() => {
            const secondarySpan = document.querySelector('.secondary-bg');
            return window.getComputedStyle(secondarySpan).getPropertyValue('background-color');
        });
        expect(backgroundColor).toBe('rgba(135, 206, 235, 0.5)');
    });
});


<!DOCTYPE html>

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Font Family Example</title>

Header Font Family

Paragraph is using a serif font.

创建一个 HTML 页面，并应用 CSS 来设置不同元素的字体族（font-family）。要求如下：

1. 页面应含有一个 <h1> 标签，类名为 header-font，内容为 "Header Font Family"，字体族需设置为 "Arial, sans-serif"。
2. 页面也包含一个 <p> 标签，类名为 paragraph-font，内容为 "Paragraph is using a serif font."，字体族需设置为 "Times New Roman, serif"。

CSS 书写要求：

1. 使用类选择器，以点号 . 开头，后跟类名来定义选择器（例如 .header-font）。
2. 在定义字体族时，首先使用具体的字体名称，然后跟一个通用的家族名称（例如 sans-serif 或 serif）作为后备选项。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, test, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Font Family Styling', () => {
    let browser;
    let page;

    beforeAll(async () => {
        browser = await puppeteer.launch({args: ['--no-sandbox']});
        page = await browser.newPage();
        await page.setContent(`#<INSERT>`);
    });

    afterAll(async () => {
        await browser.close();
    });

    test('The .header-font class should use Arial font family', async () => {
        const fontFamily = await page.evaluate(() => {
            const header = document.querySelector('.header-font');
            return window.getComputedStyle(header).getPropertyValue('font-family');
        });
        // The font-family property can vary in different browsers or systems,
        // so we check if "Arial" is included in the returned style.
        expect(fontFamily).toMatch(/Arial/);
    });

    test('The .paragraph-font class should use Times New Roman font family', async () => {
        const fontFamily = await page.evaluate(() => {
            const paragraph = document.querySelector('.paragraph-font');
            return window.getComputedStyle(paragraph).getPropertyValue('font-family');
        });
        // The font-family property can vary in different browsers or systems,
        // so we check if "Times New Roman" is included in the returned style.
        expect(fontFamily).toMatch(/Times New Roman/);
    });
});


<!DOCTYPE html>

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>CSS Exercise</title>

CSS Styling Exercise

    Hover over me

创建一个HTML页面，它包含一个标题 <h1> 元素和一个 <div> 容器。在这个 <div> 容器中包含一个 <span> 元素。请为这三个元素应用以下样式规则：

* <h1> 标题应该有一个居中的文本对齐方式，并且背景颜色应该为浅灰色。
* <div> 容器应该有一定的宽度（例如 400px），并且内部文本居中显示。除此之外，它应该具有蓝色的边框。
*<span> 元素内的文本应该是粗体的，并且在鼠标悬停时变为橙色。

请写出满足上述条件的 HTML 和 CSS 代码。
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, test, beforeAll, afterAll, it } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('CSS styling', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  afterAll(async () => {
    await browser.close();
  });

  it('<h1> element should be center aligned with a light gray background', async () => {
    const h1Styles = await page.evaluate(() => {
      const h1 = document.querySelector('h1');
      return {
        textAlign: window.getComputedStyle(h1).textAlign,
        backgroundColor: window.getComputedStyle(h1).backgroundColor
      };
    });

    expect(h1Styles.textAlign).toBe('center');
    expect(h1Styles.backgroundColor).toBe('rgb(240, 240, 240)');
  });

  it('<div> element should be 400px wide with centered text and a blue border', async () => {
    const divStyles = await page.evaluate(() => {
      const div = document.querySelector('div');
      return {
        width: window.getComputedStyle(div).width,
        textAlign: window.getComputedStyle(div).textAlign,
        borderColor: window.getComputedStyle(div).borderColor
      };
    });

    expect(divStyles.width).toBe('400px');
    expect(divStyles.textAlign).toBe('center');
    expect(divStyles.borderColor).toBe('rgb(0, 0, 255)');
  });

  it('<span> text should be bold and turn orange when hovered', async () => {
    const spanStyles = await page.evaluate(() => {
      const span = document.querySelector('span');
      return window.getComputedStyle(span).fontWeight;
    });

    expect(spanStyles).toBe('700'); // 假定 bold 相当于 font-weight 700

    await page.hover('span'); // 鼠标悬停在 span 元素上

    const colorAfterHover = await page.evaluate(() => {
      const span = document.querySelector('span');
      return window.getComputedStyle(span).color;
    });

    expect(colorAfterHover).toBe('rgb(255, 165, 0)'); // 暗示悬停后颜色应为橙色
  });
});


<!DOCTYPE html>

<meta charset="UTF-8">
<title>Text Box with Vertical Overflow</title>

创建一个名为 .text-box 的容器，它包含大量的文本。当文本超过容器的高度时，容器应该在垂直方向上显示滚动条。
容器的宽度应该是父级宽度的80%，高度固定为150px。在该容器内部，有一个 .inner-text 的子元素实际包含文本内容。
请按照这些要求编写相应的HTML和CSS代码，并确保CSS书写规范化。

CSS书写要求：

1. CSS类选择器应与HTML中的类名对应。
2. CSS应使用合理的命名，使得样式的作用一目了然。
3. CSS书写时，每个属性声明前应有一个缩进（通常是2个空格）。
4. 选择器的左大括号{放在选择器的同一行，右大括号}独占一行，并且在属性声明之间应适当留有空白行。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, it, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Text Box Overflow', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  it('should display a vertical scrollbar for overflowing content', async () => {
    const longText = Array(200).fill('Overflow test text. ').join('');

    await page.evaluate((text) => {
      document.querySelector('.inner-text').innerText = text;
    }, longText);

    const overflow = await page.evaluate(() => {
      const textBox = document.querySelector('.text-box');
      return textBox.scrollHeight > textBox.clientHeight;
    });

    expect(overflow).toBe(true);
  });

  afterAll(async () => {
    await browser.close();
  });
});


def islands(map):
    if not map:
        return 0

    def dfs(i, j):
        if i < 0 or i >= len(map) or j < 0 or j >= len(map[0]) or map[i][j] == '0':
            return
        map[i][j] = '0'  # 将访问过的陆地标记为 '0'
        dfs(i + 1, j)    # 访问下方单元格
        dfs(i - 1, j)    # 访问上方单元格
        dfs(i, j + 1)    # 访问右侧单元格
        dfs(i, j - 1)    # 访问左侧单元格

    count = 0
    for i in range(len(map)):
        for j in range(len(map[0])):
            if map[i][j] == '1':
                dfs(i, j)
                count += 1

    return count

请写一个Python函数`islands(map): int` 解决这个问题:
给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格map，计算并返回网格中岛屿的数量。岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。此外，你可以假设该网格的四条边均被水包围。
>>> islands([
  ['1','1','0','0','0'],
  ['1','1','0','0','0'],
  ['0','0','1','0','0'],
  ['0','0','0','1','1']
])
3


from functools import cmp_to_key

def cmp(p_1, p_2):

    v_1 = max(p_2[1], p_1[0] * p_1[1])
    v_2 = max(p_1[1], p_2[0] * p_2[1])

    if v_1 == v_2:
        return 0

    return -1 if v_1 < v_2 else 1

def solve(king, ministers):
    a, b = king

    ministers = sorted(ministers, key=cmp_to_key(cmp))

    ans = 0
    prod = a

    for l, r in ministers:
        ans = max(ans, prod // r)
        prod *= l

    return ans

恰逢 H 国国庆，国王邀请 n 位大臣来玩一个有奖游戏。
首先，他让每个大臣在左、右手上面分别写下一个整数，国王自己也在左、右手上各写一个整数。
然后，让这 n 位大臣排成一排，国王站在队伍的最前面。排好队后，所有的大臣都会获得国王奖赏的若干金币，每位大臣获得的金币数分别是：排在该大臣前面的所有人的左手上的数的乘积 除以 他自己右手上的数，然后向下取整得到的结果。
国王不希望某一个大臣获得特别多的奖赏，所以他想请你帮他重新安排一下队伍的顺序，使得获得奖赏最多的大臣，所获奖赏尽可能的少。注意，国王的位置始终在队伍的最前面。

用python定义函数 solve(king, ministers) ，其中king是一个二元组，代表国王左手和右手上的整数，minister是每个大臣的左手和右手上整数的二元组列表。
返回重新排列后获奖最多的大臣获奖的金币数量。


def is_black_dead(board):
    rows, cols = len(board), len(board[0])
    visited = set()

    def dfs(i, j):
        if i < 0 or i >= rows or j < 0 or j >= cols:
            return False
        if board[i][j] == 0:
            return True  # 找到空位，“气”
        if board[i][j] == 1 or (i, j) in visited:
            return False
        visited.add((i, j))
        # 检查四个方向
        return dfs(i-1, j) or dfs(i+1, j) or dfs(i, j-1) or dfs(i, j+1)

    for i in range(rows):
        for j in range(cols):
            if board[i][j] == 2 and (i, j) not in visited:
                if not dfs(i, j):
                    return True  # 找到了死亡的黑棋
    return False

用一个二维数组a表示围棋棋盘的状态，a[i][j] = 0 表示没有棋子，a[i][j] = 1 表示白棋，a[i][j] = 2 表示黑棋。给定一个棋盘状态，判断棋盘中有没有已经死了的黑子。
用python实现，定义函数 is_black_dead(a)，输入棋盘状态，输出是否有已经死了的黑子。


<!DOCTYPE html>

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Font Size Example</title>

Welcome to My Site!

This is a short introduction to my website. Enjoy your stay!

创建一个 HTML 页面，其中包含以下元素，并使用 CSS 确定其字体大小：

1. 标题元素 <h1>（类名为 title），其文本内容为 "Welcome to My Site!"，字体大小为 36px。
2. 段落 <p>（类名为 description），包含一些简短的介绍文本，字体大小为 18px。

CSS书写要求:

1. 使用类选择器根据 HTML 元素的名称进行样式定义。
2. 请使用固定像素值（例如：px）设置字体大小。
3. 确保页面的 HTML 元素与 CSS 类选择器名称相匹配。

请提供满足要求的 HTML 和 CSS 代码片段。
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, test, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Font Size Styling', () => {
    let browser;
    let page;

    beforeAll(async () => {
        browser = await puppeteer.launch({args: ['--no-sandbox']});
        page = await browser.newPage();
        await page.setContent(`#<INSERT>`);
    });

    afterAll(async () => {
        await browser.close();
    });

    test('The .title class should have a font size of 36px', async () => {
        const fontSize = await page.evaluate(() => {
            const title = document.querySelector('.title');
            return window.getComputedStyle(title).getPropertyValue('font-size');
        });
        expect(fontSize).toBe('36px');
    });

    test('The .description class should have a font size of 18px', async () => {
        const fontSize = await page.evaluate(() => {
            const description = document.querySelector('.description');
            return window.getComputedStyle(description).getPropertyValue('font-size');
        });
        expect(fontSize).toBe('18px');
    });
});


<!DOCTYPE html>

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Font Color Example</title>

这是主标题

这是一段描述文本

编写一段 HTML 代码，并利用 CSS 设置不同元素的字体颜色。要求如下：

1. 页面应该有一个 <h1> 元素，类名为 main-heading，内容为 "这是主标题"，字体颜色应该设置为蓝色（#0000FF）。
2. 页面还应该包含一个 <p> 元素，类名为 info-text，内容为 "这是一段描述文本"，字体颜色应该设为绿色（#008000）。

CSS 书写要求：

1. 使用类选择器，即以点 . 开头，后跟类名称来定义选择器（例如 .main-heading）。
2. 使用十六进制颜色代码指定元素的字体颜色。

要求在答案中返回在一起的 HTML 和 CSS 代码
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, test, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Font Color Styling', () => {
    let browser;
    let page;

    beforeAll(async () => {
        browser = await puppeteer.launch({args: ['--no-sandbox']});
        page = await browser.newPage();
        await page.setContent(`#<INSERT>`);
    });

    afterAll(async () => {
        await browser.close();
    });

    test('The .main-heading class should have a font color of blue (#0000FF)', async () => {
        const color = await page.evaluate(() => {
            const heading = document.querySelector('.main-heading');
            return window.getComputedStyle(heading).getPropertyValue('color');
        });
        expect(color).toBe('rgb(0, 0, 255)'); // 转换为 RGB 格式
    });

    test('The .info-text class should have a font color of green (#008000)', async () => {
        const color = await page.evaluate(() => {
            const text = document.querySelector('.info-text');
            return window.getComputedStyle(text).getPropertyValue('color');
        });
        expect(color).toBe('rgb(0, 128, 0)'); // 转换为 RGB 格式
    });
});


from decimal import Decimal, getcontext

def solve(R, n):
    """
    Calculate R to the power of n with high precision using the decimal module.

    :param R: A real number (0.0 < R < 99.999)
    :param n: An integer (0 < n <= 25)
    :return: R raised to the power of n
    """
    # Set the precision high enough to handle the calculation
    getcontext().prec = 50

    # Use Decimal for high precision calculation
    result = Decimal(R) ** n

    return result

对数值很大、精度很高的数进行高精度计算是一类十分常见的问题。比如，对国债进行计算就是属于这类问题。

请写一个Python函数`solve(R, n):Float` 解决这个问题:
对一个实数R( 0.0 < R < 99.999 )，要求写程序精确计算 R 的 n 次方(R^n)，其中n 是整数并且 0 < n <= 25。

>>> solve(95.123,12)
548815620517731830194541.899025343415715973535967221869852721


<!DOCTYPE html>

<meta charset="UTF-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>Centered Paragraph</title>

This paragraph is centered in the div.

请创建一个HTML页面，其中包含一个 <div> 容器。这个 <div> 容器中包含一个居中对齐的 <p> 元素，且该 <p> 元素的背景为黄色，文本颜色为黑色，并有足够的边距（padding）使得文本看起来不拥挤。
请写出满足上述条件的 HTML 和 CSS 代码。
请将答案代码用 markdown code block 格式包裹，如
```html
[code]
```

import { describe, expect, test, beforeAll, afterAll } from '@jest/globals';
import puppeteer from 'puppeteer';

describe('Centered Paragraph', () => {
  let browser;
  let page;

  beforeAll(async () => {
    browser = await puppeteer.launch({args: ['--no-sandbox']});
    page = await browser.newPage();
    await page.setContent(`#<INSERT>`);
  });

  afterAll(async () => {
    await browser.close();
  });

  it('should have a <div> that is a flex container', async () => {
    const divDisplay = await page.evaluate(() => {
      const div = document.querySelector('div');
      return window.getComputedStyle(div).display;
    });

    expect(divDisplay).toBe('flex');
  });

  it('should have a <p> element with yellow background and black text', async () => {
    const pStyles = await page.evaluate(() => {
      const p = document.querySelector('p');
      return {
        backgroundColor: window.getComputedStyle(p).backgroundColor,
        color: window.getComputedStyle(p).color,
        textAlign: window.getComputedStyle(p).textAlign,
        padding: window.getComputedStyle(p).padding
      };
    });

    expect(pStyles.backgroundColor).toBe('rgb(255, 255, 0)'); // yellow
    expect(pStyles.color).toBe('rgb(0, 0, 0)'); // black
    expect(pStyles.textAlign).toBe('center');
    expect(pStyles.padding).toBe('20px'); // assuming there's uniform padding
  });
});


def max_square(bars):
    stack = []
    max_area = 0
    bars.append(0)  # 添加一个高度为0的柱子来处理所有剩余的柱子

    for i in range(len(bars)):
        while stack and bars[stack[-1]] > bars[i]:
            height = bars[stack.pop()]
            width = i if not stack else i - stack[-1] - 1
            max_area = max(max_area, height * width)
        stack.append(i)

    return max_area

请写一个Python函数`max_square(bars): int` 解决这个问题:
一个给定的数列bars，表示一个从左向右的柱状图中每个柱子的高度，所有柱子都相邻，且宽度为 1 。计算并返回在该柱状图中，能够勾勒出来的矩形的最大面积。
>>> max_square([2,4])
4

#<INSERT>

def check():
    assert max_square([2,4]) == 4
    assert max_square([3,1,2]) == 3
    assert max_square([2,1,5,6,2,3]) == 10
check()


def first_missing_positive_num(nums):
    n = len(nums)
    for i in range(n):
        while 1 <= nums[i] <= n and nums[nums[i] - 1] != nums[i]:
            nums[nums[i] - 1], nums[i] = nums[i], nums[nums[i] - 1]

    for i in range(n):
        if nums[i] != i + 1:
            return i + 1

    return n + 1

请写一个Python函数`first_missing_positive_num(nums): int` 解决这个问题:
对于一个给定的数组nums，计算并返回其中没有出现的最小的正整数。请你实现时间复杂度为 O(n) 并且只使用常数级别额外空间的解决方案。
>>> first_missing_positive_num([1,2,0])
3

#<INSERT>

def check():
    assert first_missing_positive_num([1,2,0]) == 3
    assert first_missing_positive_num([3,4,-1,1]) == 2
    assert first_missing_positive_num([7,8,9,11,12]) == 1
    assert first_missing_positive_num([0,0,0,1,2,2,3,5]) == 4
check()


def max_subsequence(nums):
    if not nums:
        return 0

    dp = [1] * len(nums)
    for i in range(1, len(nums)):
        for j in range(i):
            if nums[j] < nums[i]:
                dp[i] = max(dp[i], dp[j] + 1)

    return max(dp)

请写一个Python函数`max_subsequence(nums): int` 解决这个问题:
对于给定的数列nums，计算并返回其中最长严格递增子序列的长度。子序列是由数组派生而来的序列，删除（或不删除）数组中的元素而不改变其余元素的顺序。例如，[3,6,2,7] 是数组 [0,3,1,6,2,2,7] 的子序列。
>>> max_subsequence([1,2,3,7,2,10])
5

#<INSERT>

def check():
    assert max_subsequence([1,2,3,7,2,10]) == 5
    assert max_subsequence([10,9,2,5,3,7,101,18]) == 4
    assert max_subsequence( [0,1,0,3,2,3]) == 4
    assert max_subsequence([7,7,7,7,7,7,7]) == 1
check()


def solve(n):
    return n > 0 and (n & (n - 1)) == 0

请写一个Python函数`solve(n): bool` 解决这个问题:
对于给定的整数n，如果n是2的幂次方。如果是，返回True ；否则，返回False 。
>>> solve(1024)
True

#<INSERT>

def check():
    assert solve(1024) == True
    assert solve(10234) == False
    assert solve(1) == True
    assert solve(2) == True
    assert solve(3) == False
    assert solve(0) == False
check()


cat linux_test_2/a.txt linux_test_2/b.txt > linux_test_2/c.txt

请用Linux 命令实现：将linux_test_2/a.txt和linux_test_2/b.txt内容合并到linux_test_2/c.txt文件中

问题：请用Linux 命令实现：给定一个文本文件 file.txt，请只打印这个文件中的第十行。

答案：
```bash
sed -n '10p' file.txt
```

----------------

问题：请用Linux 命令实现：写一个 bash 脚本以统计一个文本文件 words.txt 中每个单词出现的频率。

答案：
```bash
cat words.txt | tr -s ' ' '\n' | tr '[:upper:]' '[:lower:]' | sort | uniq -c
```

----------------

import subprocess
def init():
    import os
    os.makedirs('linux_test_2')

def check_result(model_output):
    init()
    text = """Line 1
Line 2"""
    text2 = """
Line 3
Line 4"""
    with open('linux_test_2/a.txt', 'w') as file:
        file.write(text)
    with open('linux_test_2/b.txt', 'w') as file:
        file.write(text2)
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    process = subprocess.Popen(["cat linux_test_2/c.txt"], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    output = stdout.decode()
    assert(output == "Line 1\nLine 2\nLine 3\nLine 4")
model_output = """#<INSERT>"""
check_result(model_output)


find linux_test_3 -name "*.md" > linux_test_3/md_files.txt

请用Linux 命令实现：查找linux_test_3及子目录下所有的.md文件，并将这些文件的名称输出到linux_test_3/md_files.txt中

import subprocess
def init():
    from pathlib import Path
    import os
    def create(file_name):
        path = Path(file_name)
        os.makedirs(path.parent, exist_ok=True)
        path.touch()

    for f in ['file7.md', 'file8.txt', 'folder1/file1.md', 'folder1/file2.txt',\
              'folder2/file3.md', 'folder2/file4.md', 'folder3/file6.txt', 'folder3/subfolder1/file5.md']:
        create('linux_test_3/' + f)
    return

def check_result(model_output):
    init()
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()

    process = subprocess.Popen(["cat linux_test_3/md_files.txt"], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    output = stdout.decode().strip().split('\n')
    output = sorted(output)
    expected = sorted([
        "linux_test_3/folder1/file1.md",
        "linux_test_3/folder2/file3.md",
        "linux_test_3/folder2/file4.md",
        "linux_test_3/folder3/subfolder1/file5.md",
        "linux_test_3/file7.md",
    ])
    assert(output == expected)

model_output = """#<INSERT>"""
check_result(model_output)


def convert(n):
    return int(bin(n)[2:])

请写一个Python函数`convert(n): int` 解决这个问题:
对于一个给定的十进制数n，计算并返回它的二进制。
>>> convert(1025)
10000000001

#<INSERT>

def check():
    assert convert(1) == 1
    assert convert(2) == 10
    assert convert(1024) == 10000000000
    assert convert(6342) == 1100011000110
check()


请用Linux 命令实现：将 `linux_test_12` 目录下的文件 `test_1` 改名为 `test_2`，`linux_test_12` 目录位于当前路径

import os
import subprocess
def check_result(model_output):
    os.makedirs('linux_test_12', exist_ok=True)
    with open('linux_test_12/test_1', 'w') as file:
        file.write("test")

    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

    stdout, stderr = process.communicate()

    assert(os.path.isfile('linux_test_12/test_2'))

model_output = """#<INSERT>"""

check_result(model_output)


ln -s linux_test_6/a.foo linux_test_6/b.bar

请用Linux 命令实现：给文件linux_test_6/a.foo 创建一个名为linux_test_6/b.bar的软连接

import subprocess
import os

def check_result(model_output):
    os.makedirs('linux_test_6')
    with open('linux_test_6/a.foo', 'w') as file:
        file.write("Test content")
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    assert(os.path.islink('linux_test_6/b.bar'))
    assert(os.readlink('linux_test_6/b.bar') == 'linux_test_6/a.foo')
model_output = """#<INSERT>"""
check_result(model_output)


请用Linux 命令实现：输出linux_test_1/test.txt文件中，包含"mark"的行及其上面2行的数据

import subprocess
def init():
    import os
    os.makedirs('linux_test_1', exist_ok=True)

def check_result(model_output):
    init()
    text = """Line 1
Line 2
Line 3
Line 4    mark
Line 5"""
    with open('linux_test_1/test.txt', 'w') as file:
        file.write(text)
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    output = stdout.decode()
    assert(output == "Line 2\nLine 3\nLine 4    mark\n")
model_output = """#<INSERT>"""
check_result(model_output)


请用Linux 命令实现：在 `linux_test_8` 目录下创建一个新的文件夹 `test` ，`linux_test_8` 位于当前路径

import os
def check_result(model_output):
    os.makedirs('linux_test_8')
    os.system(model_output)
    assert(os.path.isdir("linux_test_8/test"))

model_output = """#<INSERT>"""

check_result(model_output)


请用Linux 命令实现：寻找linux_test_4目录及子目录下名为test.txt的文件

import subprocess
def init():
    from pathlib import Path
    import os
    def create(file_name):
        path = Path(file_name)
        os.makedirs(path.parent, exist_ok=True)
        path.touch()

    for f in ['test.txt', 'file8.txt', 'folder1/test.txt', 'folder2/subfolder1/test.txt']:
        create('linux_test_4/' + f)
    return

def check_result(model_output):
    init()
    expected_files = ['linux_test_4/folder1/test.txt', 'linux_test_4/folder2/subfolder1/test.txt', 'linux_test_4/test.txt']
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    output_files = stdout.decode().strip().split('\n')
    assert(set(output_files) == set(expected_files))
model_output = '''#<INSERT>'''
check_result(model_output)


请用Linux 命令实现：给linux_test_5/run_test.sh加上执行权限

import subprocess
def init():
    import os
    os.makedirs('linux_test_5')

def check_result(model_output):
    init()
    with open('linux_test_5/run_test.sh', 'w') as file:
        file.write("#!/bin/bash\necho 'Script is running'")
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    process = subprocess.Popen(["ls -l linux_test_5/run_test.sh"], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    stdout, stderr = process.communicate()
    output = stdout.decode()
    assert("-rwx" in output)
model_output = """#<INSERT>"""
check_result(model_output)


请用Linux 命令实现：统计linux_test_13下的test文件中有多少行

import subprocess
def init():
    import os
    os.makedirs('linux_test_13')

def check_result(model_output):
    init()
    text = """Line 1
Line 2
Line 3
Line 4
Line 5
"""
    with open('linux_test_13/test', 'w') as file:

        file.write(text)

    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

    stdout, stderr = process.communicate()

    output = stdout.decode()
    assert(output.strip() == "5 linux_test_13/test")

model_output = """#<INSERT>"""

check_result(model_output)


请用Linux 命令实现：在linux_test_11下创建一个空的文件test

import subprocess
import os
def check_result(model_output):
    os.makedirs('linux_test_11')
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

    stdout, stderr = process.communicate()

    assert(os.path.exists('linux_test_11/test'))

model_output = """#<INSERT>"""

check_result(model_output)


请用Linux 命令实现：删除linux_test_9下面的test文件夹

#import os
#import subprocess
def check_result(model_output):
    #os.makedirs('linux_test_9/test', exist_ok=True)
    #assert os.path.exists('linux_test_9/test')

    #process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

    #stdout, stderr = process.communicate()

    #assert not os.path.exists('linux_test_9/test')

    assert model_output in set(['rm -r linux_test_9/test', 'rm -rf linux_test_9/test'])

model_output = """#<INSERT>"""

check_result(model_output)


请用Linux 命令实现：删除linux_test_10下面的test文件

#import os
#import subprocess
def check_result(model_output):
    #os.makedirs('linux_test_10', exist_ok=True)
    #with open('linux_test_10/test', 'w') as file:
    #    file.write("test")

    #process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

    #stdout, stderr = process.communicate()

    #assert(not os.path.exists('linux_test_10/test'))
    assert model_output in set(['rm linux_test_10/test', 'rm -f linux_test_10/test'])

model_output = """#<INSERT>"""

check_result(model_output)


请用Linux 命令实现：清除环境变量TEST_ENV_2的值

import subprocess
# linux_test_14
def check_result(model_output):
    # Set the environment variable
    #subprocess.run(["export TEST_ENV_2='test value'"], shell=True)

    # Run the model's command
    #subprocess.run([model_output], shell=True)

    # Check if the environment variable is unset
    #process = subprocess.Popen(["echo $TEST_ENV_2"], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    #stdout, stderr = process.communicate()
    #output = stdout.decode().strip()

    #assert(output == "")

    assert model_output in set(['unset TEST_ENV_2'])

model_output = """#<INSERT>"""

check_result(model_output)


请用Linux 命令实现：查看环境变量TEST_ENV_3的值

import os
import subprocess
def check_result(model_output):
    # linux_test_15
    os.environ['TEST_ENV_3'] = 'test_value'

    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)

    stdout, stderr = process.communicate()

    output = stdout.decode()

    assert(output == "test_value\n")

model_output = """#<INSERT>"""

check_result(model_output)


请用Linux 命令查看环境变量TEST_EVAL的值

import subprocess
def check_result(model_output):
    # linux_test_7
    expected_value = "some_test_value"
    process = subprocess.Popen(["export TEST_EVAL=some_test_value"], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
    process.communicate()
    process = subprocess.Popen([model_output], shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE, env={"TEST_EVAL": expected_value})
    stdout, stderr = process.communicate()
    output = stdout.decode().strip()
    assert(output == expected_value)
model_output = """#<INSERT>"""
check_result(model_output)


from http.server import BaseHTTPRequestHandler, HTTPServer
import json

class WebhookHandler(BaseHTTPRequestHandler):

    def do_POST(self):
        if self.path == '/webhook':
            # 获取内容长度
            content_length = int(self.headers['Content-Length'])
            # 读取请求体数据
            post_data = self.rfile.read(content_length)

            # 尝试解析JSON数据
            try:
                data = json.loads(post_data)
                # 打印收到的信息
                print("Received webhook:")
                print(f"Message:{data['message']}")
                print(f"Task ID: {data['taskId']}")
            except json.JSONDecodeError:
                # 如果不是JSON格式，就打印错误信息
                print("Received non-JSON payload")
                self.send_response(400)  # Bad Request
                self.end_headers()
                return

            #送响应
            self.send_response(200)  # OK
            self.send_header('Content-type', 'application/json')
            self.end_headers()
            response = {"status": "received"}
            self.wfile.write(json.dumps(response).encode('utf-8'))
        else:
            # 如果路径不是/webhook，则返回404
            self.send_response(404)  # Not Found
            self.end_headers()

def run(server_class=HTTPServer, handler_class=WebhookHandler, port=3410):
    server_address = ('', port)
    httpd = server_class(server_address, handler_class)
    print(f'Starting httpd server on port {port}...')
    httpd.serve_forever()

if __name__ == "__main__":
    run()

在许多异步通信场景中，例如在服务器端处理完一个耗时任务后，使用HTTP回调（也称为Webhook）通知客户端是一种常见做法。在本题中，你需要设计并实现一个简单的HTTP回调处理器。

创建一个Python脚本，实现以下功能：

1. 启动一个简单的HTTP服务器，监听3410端口。
2. 服务器应该能处理POST请求，路径为`/webhook`。
3. 当接收到POST请求时，服务器应该从请求体中提取JSON格式的数据，并打印出来。
4. 收到请求后，服务器应该返回状态码200，并且响应体为`{"status": "received"}`。

假设收到的JSON格式如下：

```json
{
  "message": "Task completed",
  "taskId": "12345"
}
```

你需要打印出类似于以下的信息：

```plaintext
Received webhook:
Message: Task completed
Task ID: 12345
```

请把代码写在一个markdown ```python```包裹中

def check(model_output):
    import requests
    import subprocess
    import time
    import os
    import signal
    import shlex
    import json
    from requests.auth import HTTPBasicAuth
    os.makedirs('http_test_1', exist_ok=True)
    with open("http_test_1/app.py", "w") as app_file:
        app_file.write(model_output)
    cmd = shlex.split("python3 app.py")
    pro = subprocess.Popen(cmd, stdout=subprocess.PIPE,
                        preexec_fn=os.setsid,cwd = "http_test_1/")

    time.sleep(5)

    try:

        # 定义服务器的URL
        url = 'http://localhost:3410/webhook'

        # 定义要发送的数据
        data = {
        'message': 'Task completed',
        'taskId': '12345'
        }

        # 将数据转换为JSON字符串
        json_data = json.dumps(data)

        # 发送POST请求
        response = requests.post(url, data=json_data, headers={'Content-Type': 'application/json'})

        # 断言响应状态码为200
        assert response.status_code == 200, f"Expected status code 200, but got {response.status_code}"

        # 断言响应内容
        expected_response = {"status": "received"}
        assert response.json() == expected_response, f"Expected response {expected_response}, but got {response.json()}"

        # 测试完成，如果没有断言错误，则输出成功信息
        print("Webhook handler passed the test successfully.")
    except AssertionError as e:
        raise AssertionError("Model output does not match")
    finally:
        os.killpg(os.getpgid(pro.pid), signal.SIGTERM)
        time.sleep(1)
        # os.remove("http_test_1/app.py")
model_output = """#<INSERT>"""
check(model_output)


from typing import List

def equal_row_and_column_pairs(grid: List[List[int]]) -> int:
    """
    Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.
    A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).

    Args:
    grid (List[List[int]]): A 0-indexed n x n integer matrix.

    Returns:
    int: The number of equal row and column pairs.
    Example:
    assert equal_row_and_column_pairs([[3,2,1],[1,7,6],[2,7,7]]) == 1
    """

    n = len(grid)
    count = 0

    # Precompute column vectors and store them in a dictionary
    column_vectors = {j: [grid[i][j] for i in range(n)] for j in range(n)}

    for i in range(n):
        for j in range(n):
            if grid[i] == column_vectors[j]:
                count += 1

    return count

重新实现以下代码以优化其性能。
```python
from typing import List

def equal_row_and_column_pairs(grid: List[List[int]]) -> int:
    '''
    Given a 0-indexed n x n integer matrix grid, return the number of pairs (ri, cj) such that row ri and column cj are equal.
    A row and column pair is considered equal if they contain the same elements in the same order (i.e., an equal array).

    Args:
    grid (List[List[int]]): A 0-indexed n x n integer matrix.

    Returns:
    int: The number of equal row and column pairs.
    Example:
    assert equal_row_and_column_pairs([[3,2,1],[1,7,6],[2,7,7]]) == 1
    '''

    n = len(grid)
    count = 0
    for i in range(n):
        for j in range(n):
            if grid[i] == [grid[k][j] for k in range(n)]:
                count += 1
    return count
```

#<INSERT>

from typing import List

def baseline(grid: List[List[int]]) -> int:
    n = len(grid)
    count = 0
    for i in range(n):
        for j in range(n):
            if grid[i] == [grid[k][j] for k in range(n)]:
                count += 1
    return count

assert equal_row_and_column_pairs([[3,2,1],[1,7,6],[2,7,7]]) == 1
assert equal_row_and_column_pairs([[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]) == 3

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
test_grid = [[3,1,2,2],[1,4,4,5],[2,4,2,2],[2,4,2,2]]
args = [test_grid]
assert _is_outperform(baseline, equal_row_and_column_pairs, args)


def longest_substring_without_repeating_chars(s: str) -> int:
    """
    Find the length of the longest substring without repeating characters in the given string.

    Args:
    s (str): The input string, consists of English letters, digits, symbols, and spaces.

    Returns:
    int: The length of the longest substring without repeating characters.
    Example:
    assert longest_substring_without_repeating_chars("abcabcbb") == 3
    """

    n = len(s)
    chars = set()
    res = 0
    left, right = 0, 0

    while right < n:
        if s[right] not in chars:
            chars.add(s[right])
            res = max(res, right - left + 1)
            right += 1
        else:
            chars.remove(s[left])
            left += 1

    return res

重新实现以下代码以优化其性能。
```python
def longest_substring_without_repeating_chars(s: str) -> int:
    '''
    Find the length of the longest substring without repeating characters in the given string.

    Args:
    s (str): The input string, consists of English letters, digits, symbols, and spaces.

    Returns:
    int: The length of the longest substring without repeating characters.
    Example:
    assert longest_substring_without_repeating_chars("abcabcbb") == 3
    '''

    def all_unique(substr: str) -> bool:
        chars = set()
        for char in substr:
            if char in chars:
                return False
            chars.add(char)
        return True

    res = 0
    for i in range(len(s)):
        for j in range(i + 1, len(s) + 1):
            if all_unique(s[i:j]):
                res = max(res, j - i)
    return res
```

#<INSERT>

def baseline(s: str) -> int:
    def all_unique(substr: str) -> bool:
        chars = set()
        for char in substr:
            if char in chars:
                return False
            chars.add(char)
        return True

    res = 0
    for i in range(len(s)):
        for j in range(i + 1, len(s) + 1):
            if all_unique(s[i:j]):
                res = max(res, j - i)
    return res

assert longest_substring_without_repeating_chars("abcabcbb") == 3
assert longest_substring_without_repeating_chars("bbbbb") == 1
assert longest_substring_without_repeating_chars("pwwkew") == 3

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = ["abcabcbb"]
assert _is_outperform(baseline, longest_substring_without_repeating_chars, args)


from flask import Flask, request, Response
from functools import wraps

app = Flask(__name__)

users = {
    "username": "password",
}

def authenticate():
    '''Sends a 401 response that enables basic auth'''
    return Response(
    'Could not verify your access level for that URL.'
    'You have to login with proper credentials', 401,
    {'WWW-Authenticate': 'Basic realm="Login Required"'})

def requires_auth(f):
    @wraps(f)
    def decorated(*args, **kwargs):
        auth = request.authorization
        if not auth or not check_auth(auth.username, auth.password):
            return authenticate()
        return f(*args, **kwargs)
    return decorated

def check_auth(username, password):
    if username in users and users[username] == password:
        return True
    return False

@app.route('/')
@requires_auth
def hello_world():
    return 'Hello, World!'

if __name__ == '__main__':
    app.run()

实现基于Basic Authentication的用户鉴权机制。使用Python Flask框架创建一个简单的Web应用，要求如下：

- 应用中有一个已预定义好的用户字典，模拟用户数据存储。
- 定义一个装饰器 `requires_auth`，用于对HTTP请求进行鉴权操作，如果鉴权成功则返回被装饰函数的执行结果，如果鉴权失败则返回带有 401 状态码的 HTTP 响应，并在响应头部 WWW-Authenticate 字段中提示需要 Basic Authentication。
- 定义一个 `check_auth` 函数，用于核对请求中Basic Authentication的用户名和密码是否与预定义的用户字典中的数据相匹配。
- 创建一个 `'/'` 路由，对应一个返回 'Hello, World!' 的函数，并且此函数使用了 `requires_auth` 装饰器，也就是说访问此路由需要进行Basic Authentication鉴权。

完成后的应用能够处理对 `'/'` 路由的 GET 请求，对提供了正确 Basic Authentication 信息的请求返回 'Hello, World!'，并对提供了错误或无 Basic Authentication 信息的请求返回 401 状态码和错误信息。
自定义的用户字典如下：
users = {
    "username": "password",
}
请把代码写在一个markdown ```python```包裹中

def check(model_output):
    import requests
    import subprocess
    import time
    import os
    import signal
    import shlex
    from requests.auth import HTTPBasicAuth
    os.makedirs('auth_test_3', exist_ok=True)
    with open("auth_test_3/app.py", "w") as app_file:
        app_file.write(model_output)
    cmd = shlex.split("flask run -h localhost -p 3401")
    pro = subprocess.Popen(cmd, stdout=subprocess.PIPE,
                        preexec_fn=os.setsid,cwd = "auth_test_3/")

    time.sleep(5)

    try:
        response = requests.get('http://localhost:3401/', auth=HTTPBasicAuth('username', 'password'))
        assert response.status_code == 200, "Status code is not 200"
        response = requests.get('http://localhost:3401/')
        assert response.status_code == 401, "Status code is not 401"
    except AssertionError as e:
        raise AssertionError("Model output does not match")
    finally:
        os.killpg(os.getpgid(pro.pid), signal.SIGTERM)
        time.sleep(1)
        os.remove("auth_test_3/app.py")
model_output ="""#<INSERT>"""
check(model_output)


from typing import List

def search_insert_position(nums: List[int], target: int) -> int:
    """
    Find the index of the target value in a sorted array of distinct integers. If the target is not found,
    return the index where it would be if it were inserted in order.

    Args:
    nums (List[int]): A sorted list of distinct integers.
    target (int): The target value to find or insert.

    Returns:
    int: The index of the target value in the array or the index where it would be inserted.
    Example:
    assert search_insert_position([1, 3, 5, 6], 5) == 2
    """

    left, right = 0, len(nums) - 1

    while left <= right:
        mid = (left + right) // 2
        if nums[mid] == target:
            return mid
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1

    return left

重新实现以下代码以优化其性能。
```python
from typing import List

def search_insert_position(nums: List[int], target: int) -> int:
    '''
    Find the index of the target value in a sorted array of distinct integers. If the target is not found,
    return the index where it would be if it were inserted in order.

    Args:
    nums (List[int]): A sorted list of distinct integers.
    target (int): The target value to find or insert.

    Returns:
    int: The index of the target value in the array or the index where it would be inserted.
    Example:
    assert search_insert_position([1, 3, 5, 6], 5) == 2
    '''

    for i in range(len(nums)):
        if nums[i] >= target:
            return i
    return len(nums)
```

#<INSERT>

from typing import List
def baseline(nums: List[int], target: int) -> int:
    for i in range(len(nums)):
        if nums[i] >= target:
            return i
    return len(nums)

assert search_insert_position([1, 3, 5, 6], 5) == 2
assert search_insert_position([1, 3, 5, 6], 2) == 1
assert search_insert_position([1, 3, 5, 6], 7) == 4
assert search_insert_position([1, 2, 3, 4, 5, 6, 9, 10], 7) == 6

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
import random
arr = sorted([random.randint(1, 5000) for _ in range(5000)])
args = [arr, 876]
assert _is_outperform(baseline, search_insert_position, args)


import queue
import threading
import time

def worker(q, n, result_list):
    while True:
        item = q.get()
        if item is None:
            break
        result_list.append(item)
        time.sleep(1)  # simulate a long process
        # Indicate that this task is done
        q.task_done()

def threaded_task(num_worker_threads, source):
    q = queue.Queue()
    threads = []
    result_list = []

    for i in range(num_worker_threads):
        t = threading.Thread(target=worker, args=(q, i, result_list))
        t.start()
        threads.append(t)

    # fill the queue with source items
    for item in source:
        q.put(item)

    q.join()  # block until all tasks are done

    # stop workers
    for i in range(num_worker_threads):
        q.put(None)
    for t in threads:
        t.join()

    return result_list

用Python实现一个多线程任务处理机制。所要求的功能如下：

- 创建一个可以接收任务的消息队列。
- 创建多个工作线程，每个线程都可以从队列中获取任务进行处理。模拟每个任务的处理时间为1秒。
- 当所有任务都被处理完后，停止工作线程并返回所有已处理的任务。

你需要完成一个函数 `threaded_task`，其输入参数为工作线程的数量 `num_worker_threads` 和一个来源于待处理任务的列表 `source`，输出为一个包含所有已处理过任务的列表。

这个函数的主要工作流程为：

- 首先，它会初始化一个空的队列 `q`，线程列表 `threads` 和结果列表 `result_list`。
- 然后，它会根据 `num_worker_threads` 的大小创建相应数量的工作线程并启动。这些工作线程会从队列中取出任务并处理，处理结果会添加到 `result_list` 中。
- 之后，它将所有来源于 `source` 的任务添加到队列中。
- 之后，它会等待队列中所有的任务都被处理完。
- 最后，它会停止所有的工作线程并返回 `result_list`，也就是所有已处理的任务。

工作线程的工作流程为：

- 不断地从队列中获取任务来处理，如果获取到的任务为 `None`，那就结束循环并退出线程。
- 把处理的结果添加到 `result_list` 中，模拟每个任务的处理时间为1秒。
- 每个任务处理完后，调用 `q.task_done()` 来告诉队列此任务已经处理完毕。
请把代码写在一个markdown ```python```包裹中

#<INSERT>

def check(threaded_task):
    result = threaded_task(2, range(5))
    assert len(result) == 5,"Model output incorrect"
check(threaded_task)


from nameko.rpc import rpc, RpcProxy

class MovieService:
    name = "movie_service"

    @rpc
    def add_movie(self, movie_name):
        with open('nameko_test_1/movies.txt', 'a') as file:
            file.write(f'{movie_name}\n')
        return "Movie added!"

    @rpc
    def get_movies(self):
        with open('nameko_test_1/movies.txt', 'r') as file:
            result = file.readlines()
            result = [movie.strip("\n") for movie in result]
        return result

    @rpc
    def remove_movie(self, movie_name):
        with open('nameko_test_1/movies.txt', 'r') as file:
            movies = file.readlines()

        with open('nameko_test_1/movies.txt', 'w') as file:
            for movie in movies:
                if movie.strip("\n") != movie_name:
                    file.write(f"{movie}")

        return "Movie removed!"

使用Nameko框架实现一个电影服务

你需要使用Nameko框架创建一个微服务，这个服务用于对一个电影清单进行增加和删除电影、获取所有电影的操作。

你需要创建一个名为 `MovieService` 的类，这个类定义了一个微服务，该服务的名称为 "movie_service"。这个类需要包含以下的RPC方法：

- `add_movie`: 此方法接收一个电影名称（字符串）作为参数，将电影名称添加到 `nameko_test_1/movies.txt` 文件的尾部。每个电影名占一行。方法完成后，返回字符串 "Movie added!"。

- `get_movies`: 此方法不需要参数。它将读取 `nameko_test_1/movies.txt` 文件中的所有电影名，并以列表形式返回。每一个电影名（字符串）是返回列表的一个元素。

- `remove_movie`: 此方法接收一个电影名称（字符串）作为参数，从 `nameko_test_1/movies.txt` 文件中移除该名称。完成后，返回字符串 "Movie removed!"。

注意，在 `remove_movie` 方法中，你需要先读取所有的电影名称到内存中，并且在删除相应的电影后，需要将剩余的电影列表重新写回到文件中。

请注意，文件操作需要阻塞执行，所以可能不适用于真正的微服务环境，这里我们只是用来进行练习和说明问题。在实际环境中，你可能会使用数据库或其他非阻塞的IO操作来替代文件操作。
请把代码写在一个markdown ```python```包裹中

#<INSERT>

def check(MovieService):
    from nameko.testing.services import worker_factory
    import os
    os.makedirs('nameko_test_1', exist_ok=True)
    try:
        service = worker_factory(MovieService)
        service.add_movie("Interstellar")
        movies = service.get_movies()
        print(movies)
        assert movies==["Interstellar"]
        service.remove_movie("Interstellar")
        movies = service.get_movies()
        assert movies==[]
    except AssertionError as e:
        raise AssertionError("Model output does not match")
    finally:
        os.remove("nameko_test_1/movies.txt")

check(MovieService)


class CircularQueue:
    def __init__(self, capacity: int):
        self.queue = [None] * capacity
        self.capacity = capacity
        self.head = 0
        self.tail = -1
        self.size = 0

    def enqueue(self, value: int) -> bool:
        if self.is_full():
            return False
        self.tail = (self.tail + 1) % self.capacity
        self.queue[self.tail] = value
        self.size += 1
        return True

    def dequeue(self) -> bool:
        if self.is_empty():
            return False
        self.queue[self.head] = None
        self.head = (self.head + 1) % self.capacity
        self.size -= 1
        return True

    def front(self) -> int:
        if self.is_empty():
            return -1
        return self.queue[self.head]

    def rear(self) -> int:
        if self.is_empty():
            return -1
        return self.queue[self.tail]

    def is_empty(self) -> bool:
        return self.size == 0

    def is_full(self) -> bool:
        return self.size == self.capacity

你需要使用 Python 实现一个固定大小的循环队列类 CircularQueue。这个类应该支持以下操作：
构造函数 __init__(self, capacity: int) - 初始化队列，指定队列的最大容量。
enqueue(self, value: int) -> bool - 向队列插入一个元素。如果操作成功，返回 True；如果队列已满，返回 False。
dequeue(self) -> bool - 从队列中删除一个元素。如果操作成功，返回 True；如果队列为空，返回 False。
front(self) -> int - 获取队列首元素。如果队列为空，返回 -1。
rear(self) -> int - 获取队列尾元素。如果队列为空，返回 -1。
is_empty(self) -> bool - 检查队列是否为空。
is_full(self) -> bool - 检查队列是否已满。请把代码写在一个markdown ```python```包裹中

#<INSERT>
def check():
    cq = CircularQueue(3)
    assert cq.is_empty() == True
    assert cq.is_full() == False
    assert cq.enqueue(1) == True
    assert cq.enqueue(2) == True
    assert cq.front() == 1
    assert cq.rear() == 2
    assert cq.enqueue(3) == True
    assert cq.is_full() == True
    assert cq.enqueue(4) == False
    assert cq.dequeue() == True
    assert cq.is_full() == False
    assert cq.front() == 2
    assert cq.dequeue() == True
    assert cq.dequeue() == True
    assert cq.dequeue() == False
    assert cq.is_empty() == True
    assert cq.front() == -1
    assert cq.rear() == -1
check()


def can_rotate_string(s: str, goal: str) -> bool:
    """
    Determine if the string s can become the string goal after some number of shifts.

    A shift on s consists of moving the leftmost character of s to the rightmost position.

    Args:
    s (str): The initial string, consisting of lowercase English letters.
    goal (str): The target string, consisting of lowercase English letters.

    Returns:
    bool: True if s can become goal after some number of shifts, False otherwise.
    Example:
    assert can_rotate_string("abcde", "cdeab") == True
    """

    if len(s) != len(goal):
        return False

    return goal in (s * 2)

重新实现以下代码以优化其性能。
```python
def can_rotate_string(s: str, goal: str) -> bool:
    '''
    Determine if the string s can become the string goal after some number of shifts.

    A shift on s consists of moving the leftmost character of s to the rightmost position.

    Args:
    s (str): The initial string, consisting of lowercase English letters.
    goal (str): The target string, consisting of lowercase English letters.

    Returns:
    bool: True if s can become goal after some number of shifts, False otherwise.
    Example:
    assert can_rotate_string("abcde", "cdeab") == True
    '''

    if len(s) != len(goal):
        return False
    for i in range(len(s)):
        if s[i:] + s[:i] == goal:
            return True
    return False
```

#<INSERT>

def baseline(s: str, goal: str) -> bool:
    if len(s) != len(goal):
        return False
    for i in range(len(s)):
        if s[i:] + s[:i] == goal:
            return True
    return False

assert can_rotate_string("abcde", "cdeab") == True
assert can_rotate_string("abcde", "abced") == False

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = ["abcde", "cdeab"]
assert _is_outperform(baseline, can_rotate_string, args)


from typing import List

def two_sum_sorted(numbers: List[int], target: int) -> List[int]:
    left, right = 0, len(numbers) - 1

    while left < right:
        current_sum = numbers[left] + numbers[right]
        if current_sum == target:
            return [left + 1, right + 1]
        elif current_sum < target:
            left += 1
        else:
            right -= 1

    return []

重新实现以下代码以优化其性能。
```python
from typing import List

def two_sum_sorted(numbers: List[int], target: int) -> List[int]:
    '''
    Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order,
    find two numbers such that they add up to a specific target number.
    The function returns the indices of the two numbers, added by one as an integer array [index1, index2] of length 2.

    Args:
    numbers (List[int]): A sorted list of integers in non-decreasing order.
    target (int): The target sum of the two numbers in the list.

    Returns:
    List[int]: A list containing the indices of the two numbers, index1 and index2, added by one.
    Example:
    assert two_sum_sorted([2,7,11,15], 9) == [1, 2]
    '''

    for i in range(len(numbers)):
        for j in range(i+1, len(numbers)):
            if numbers[i] + numbers[j] == target:
                return [i+1, j+1]
    return []
```

#<INSERT>

from typing import List

def baseline(numbers: List[int], target: int) -> List[int]:
    for i in range(len(numbers)):
        for j in range(i+1, len(numbers)):
            if numbers[i] + numbers[j] == target:
                return [i+1, j+1]
    return []

assert two_sum_sorted([2,7,11,15], 9) == [1, 2]
assert two_sum_sorted([2,3,4], 6) == [1, 3]
assert two_sum_sorted([-1,0], -1) == [1, 2]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[2,7,8,10,11,15], 13]
assert _is_outperform(baseline, two_sum_sorted, args)


from typing import List

def increasing_triplet_subsequence(nums: List[int]) -> bool:
    """
    Determines if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k].

    Args:
    nums (List[int]): A list of integers.

    Returns:
    bool: True if there exists a triplet subsequence, False otherwise.
    Example:
    assert increasing_triplet_subsequence([1, 2, 3, 4, 5]) == True
    """

    first = second = float('inf')

    for num in nums:
        if num <= first:
            first = num
        elif num <= second:
            second = num
        else:
            return True

    return False

重新实现以下代码以优化其性能。
```python
from typing import List

def increasing_triplet_subsequence(nums: List[int]) -> bool:
    '''
    Determines if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k].

    Args:
    nums (List[int]): A list of integers.

    Returns:
    bool: True if there exists a triplet subsequence, False otherwise.
    Example:
    assert increasing_triplet_subsequence([1, 2, 3, 4, 5]) == True
    '''

    n = len(nums)
    for i in range(n):
        for j in range(i+1, n):
            for k in range(j+1, n):
                if nums[i] < nums[j] < nums[k]:
                    return True
    return False
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> bool:
    n = len(nums)
    for i in range(n):
        for j in range(i+1, n):
            for k in range(j+1, n):
                if nums[i] < nums[j] < nums[k]:
                    return True
    return False

assert increasing_triplet_subsequence([1, 2, 3, 4, 5]) == True
assert increasing_triplet_subsequence([5, 4, 3, 2, 1]) == False
assert increasing_triplet_subsequence([2, 1, 5, 0, 4, 6]) == True

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[2, 1, 5, 0, 4, 6]]
assert _is_outperform(baseline, increasing_triplet_subsequence, args)


from typing import List

def find_repeated_element(nums: List[int]) -> int:
    """
    Find the element that is repeated n times in a 2n-sized array.

    Args:
    nums (List[int]): An integer array with 2n elements, containing n+1 unique elements and one of them is repeated exactly n times.

    Returns:
    int: The element that is repeated n times.
    Example:
    assert find_repeated_element([1, 2, 3, 3]) == 3
    """

    seen = set()
    for num in nums:
        if num in seen:
            return num
        seen.add(num)

重新实现以下代码以优化其性能。
```python
from collections import Counter

def find_repeated_element(nums):
    '''
    Find the element that is repeated n times in a 2n-sized array.

    Args:
    nums (List[int]): An integer array with 2n elements, containing n+1 unique elements and one of them is repeated exactly n times.

    Returns:
    int: The element that is repeated n times.
    Example:
    assert find_repeated_element([1, 2, 3, 3]) == 3
    '''

    count = {}
    for num in nums:
        if num in count:
            count[num] += 1
        else:
            count[num] = 1

    for num, freq in count.items():
        if freq == len(nums) // 2:
            return num
```

#<INSERT>

from collections import Counter

def baseline(nums):
    count = {}
    for num in nums:
        if num in count:
            count[num] += 1
        else:
            count[num] = 1

    for num, freq in count.items():
        if freq == len(nums) // 2:
            return num

assert find_repeated_element([1, 2, 3, 3]) == 3
assert find_repeated_element([2, 1, 2, 5, 3, 2]) == 2
assert find_repeated_element([5, 1, 5, 2, 5, 3, 5, 4]) == 5

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    print(f'p={p}, {sum(rt0)}, {sum(rt1)}')
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[5, 1, 5, 2, 5, 3, 5, 4]]
assert _is_outperform(baseline, find_repeated_element, args, samples=100)


def is_palindrome_number(x: int) -> bool:
    """
    Determine if the given integer is a palindrome number.

    A palindrome number is a number that remains the same when its digits are reversed.

    Args:
    x (int): The integer to check if it's a palindrome number.

    Returns:
    bool: True if the integer is a palindrome number, False otherwise.
    Example:
    assert is_palindrome_number(121) == True
    """

    if x < 0 or (x % 10 == 0 and x != 0):
        return False

    reversed_number = 0
    while x > reversed_number:
        reversed_number = reversed_number * 10 + x % 10
        x //= 10

    return x == reversed_number or x == reversed_number // 10

重新实现以下代码以优化其性能。
```python
def is_palindrome_number(x: int) -> bool:
    '''
    Determine if the given integer is a palindrome number.

    A palindrome number is a number that remains the same when its digits are reversed.

    Args:
    x (int): The integer to check if it's a palindrome number.

    Returns:
    bool: True if the integer is a palindrome number, False otherwise.
    Example:
    assert is_palindrome_number(121) == True
    '''

    s = str(x)
    return s == s[::-1]
```

#<INSERT>

def baseline(x: int) -> bool:
    s = str(x)
    return s == s[::-1]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

assert is_palindrome_number(121) == True
assert is_palindrome_number(-121) == False
assert is_palindrome_number(10) == False
assert is_palindrome_number(11111111111111111111111111111111111111111111111111111111110) == False

# compare
args = [-11111111111111111111111111111111111111111111111111111111110]
assert _is_outperform(baseline, is_palindrome_number, args)


from concurrent import futures
import grpc

import grpc_test_1.example_pb2 as example_pb2
import grpc_test_1.example_pb2_grpc as example_pb2_grpc

class Greeter(example_pb2_grpc.GreetServicer):
    def Greeting(self, request, context):
        return example_pb2.GreetResponse(message=f"Hello {request.name}!")

def serve():
    server = grpc.server(futures.ThreadPoolExecutor(max_workers=10))
    example_pb2_grpc.add_GreetServicer_to_server(Greeter(), server)
    server.add_insecure_port('[::]:3405')
    server.start()
    server.wait_for_termination()

if __name__ == "__main__":
    serve()

请用Python实现一个基于gRPC的简单服务器。这个服务器应该提供一个名为 Greet 的服务，该服务包含一个方法 Greeting。此方法接收一个 name 参数，并返回一个格式为 "Hello {name}!" 的问候语消息。

服务的定义应该基于已有example.proto 文件，这个文件定义了 Greet 服务以及 GreetRequest 和 GreetResponse 消息。其中grpc_test_1/example_pb2.proto文件的内容如下：：
syntax = "proto3";

service Greet {
  rpc Greeting (GreetRequest) returns (GreetResponse);
}

message GreetRequest {
  string name = 1;
}

message GreetResponse {
  string message = 1;
}

你的任务是使用 example_pb2 和 example_pb2_grpc 这两个由 grpc_tools 生成的模块（已有）来实现服务器的具体逻辑。服务器应该能在本地的3405端口上接收请求，并且能够处理并发请求。
请把代码写在一个markdown ```python```包裹中


from typing import List

def max_product_subarray(nums: List[int]) -> int:
    if not nums:
        return 0

    max_product = min_product = res = nums[0]

    for i in range(1, len(nums)):
        if nums[i] < 0:
            max_product, min_product = min_product, max_product

        max_product = max(nums[i], max_product * nums[i])
        min_product = min(nums[i], min_product * nums[i])

        res = max(res, max_product)

    return res

重新实现以下代码以优化其性能。
```python
from typing import List

def max_product_subarray(nums: List[int]) -> int:
    '''
    Find a subarray within the given integer array that has the largest product and return the product.

    Args:
    nums (List[int]): A list of integers.

    Returns:
    int: The product of the subarray with the largest product.
    Example:
    assert max_product_subarray([2, 3, -2, 4]) == 6
    '''

    n = len(nums)
    max_product = float('-inf')

    for i in range(n):
        for j in range(i, n):
            product = 1
            for k in range(i, j + 1):
                product *= nums[k]
            max_product = max(max_product, product)

    return max_product
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> int:
    n = len(nums)
    max_product = float('-inf')

    for i in range(n):
        for j in range(i, n):
            product = 1
            for k in range(i, j + 1):
                product *= nums[k]
            max_product = max(max_product, product)

    return max_product

assert max_product_subarray([2, 3, -2, 4]) == 6
assert max_product_subarray([-2, 0, -1]) == 0

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[2, 3, -2, 4]]
assert _is_outperform(baseline, max_product_subarray, args)


from typing import List

def sort_array_by_parity_II(nums: List[int]) -> List[int]:
    n = len(nums)
    j = 1
    for i in range(0, n, 2):
        if nums[i] % 2:
            while nums[j] % 2:
                j += 2
            nums[i], nums[j] = nums[j], nums[i]
    return nums

重新实现以下代码以优化其性能。
```python
from typing import List

def sort_array_by_parity_II(nums: List[int]) -> List[int]:
    '''
    Sort an array of integers such that odd numbers are at odd indices and even numbers are at even indices.

    Args:
    nums (List[int]): The input array of integers where half of the integers are odd and the other half are even.

    Returns:
    List[int]: The sorted array that satisfies the condition.
    Example:
    assert sort_array_by_parity_II([4, 2, 5, 7]) == [4, 5, 2, 7]
    '''

    res = [0] * len(nums)
    even_index = 0
    odd_index = 1

    for num in nums:
        if num % 2 == 0:
            res[even_index] = num
            even_index += 2
        else:
            res[odd_index] = num
            odd_index += 2

    return res
```

#<INSERT>

from typing import List

def baseline(nums: List[int]) -> List[int]:
    res = [0] * len(nums)
    even_index = 0
    odd_index = 1

    for num in nums:
        if num % 2 == 0:
            res[even_index] = num
            even_index += 2
        else:
            res[odd_index] = num
            odd_index += 2

    return res

assert sort_array_by_parity_II([4, 2, 5, 7]) == [4, 5, 2, 7]
assert sort_array_by_parity_II([2, 3]) == [2, 3]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    print(f'p={p}, {sum(rt0)}, {sum(rt1)}')
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
args = [[4, 2, 5, 7]]
assert _is_outperform(baseline, sort_array_by_parity_II, args, repeats=2000, samples=100)


from collections import Counter
from typing import List

def intersect(nums1: List[int], nums2: List[int]) -> List[int]:
    res = []
    counter1 = {}
    for num in nums1:
        counter1[num] = counter1.get(num, 0) + 1

    for num in nums2:
        if num in counter1 and counter1[num] > 0:
            res.append(num)
            counter1[num] -= 1

    return res

重新实现以下代码以优化其性能。
```python
from collections import Counter

def intersect(nums1, nums2):
    '''
    Find the intersection of two integer arrays nums1 and nums2. Each element in the result must
    appear as many times as it shows in both arrays. The result can be returned in any order.

    Args:
    nums1 (List[int]): The first list of integers.
    nums2 (List[int]): The second list of integers.

    Returns:
    List[int]: The list containing the intersection of nums1 and nums2.
    Example:
    assert intersect([1, 2, 2, 1], [2, 2]) == [2, 2]
    '''

    res = []
    for num in nums1:
        if num in nums2:
            res.append(num)
            nums2.remove(num)
    return res
```

#<INSERT>

from collections import Counter
from typing import List

def baseline(nums1, nums2):
    res = []
    for num in nums1:
        if num in nums2:
            res.append(num)
            nums2.remove(num)
    return res

assert intersect([1, 2, 2, 1], [2, 2]) == [2, 2]
assert intersect([1, 2, 2, 1], [2]) == [2]
assert intersect([4, 9, 5], [9, 4, 9, 8, 4]) == [9, 4]

from typing import Callable, Iterable, Any
def _is_outperform(basefunc: Callable, targetfunc: Callable, args: Iterable[Any], repeats: int = 1000, samples: int = 10):
    import timeit
    from scipy.stats import ttest_ind
    # Warmup, drop result.
    _ = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]
    rt0 = [timeit.timeit(lambda: targetfunc(*args), number=repeats) for _ in range(samples)]
    rt1 = [timeit.timeit(lambda: basefunc(*args), number=repeats) for _ in range(samples)]

    # Test rt0/rt1 have identical average values with p 0.05.
    stat, p = ttest_ind(rt0, rt1, equal_var=True)
    if p > 0.05:
        return False
    return sum(rt0) < sum(rt1)

#compare
import random
test_array = [1, 3, 4, 5, 9, 10, 12, 1999, 1]
temp = list(range(2000, 5000))
random.shuffle(temp)
test_array += temp
test_array += [2, 2]
test_array += temp
test_array2 = list(range(2000, 9000))
random.shuffle(test_array2)
args = [test_array, test_array2]
assert _is_outperform(baseline, intersect, args, repeats=10, samples=5)


import unittest

class TestHasPairWithSum(unittest.TestCase):
    def test_empty_list(self):
        self.assertFalse(has_pair_with_sum([], 10), "Empty list should return False")

    def test_single_element(self):
        self.assertFalse(has_pair_with_sum([5], 10), "Single element list should return False")

    def test_no_pair(self):
        self.assertFalse(has_pair_with_sum([1, 2, 3, 9], 8), "Should return False if no pair adds up to target_sum")

    def test_with_pair(self):
        self.assertTrue(has_pair_with_sum([1, 2, 4, 4], 8), "Should return True if a pair adds up to target_sum")

    def test_negative_numbers(self):
        self.assertTrue(has_pair_with_sum([-1, -2, -4, -4, 8], 4), "Should handle negative numbers correctly")

给has_pair_with_sum 函数写测试用例, has_pair_with_sum函数如下：
```python
def has_pair_with_sum(number_list, target_sum):
    seen = set()
    for number in number_list:
        if target_sum - number in seen:
            return True
        seen.add(number)
    return False
```

import unittest

def has_pair_with_sum(number_list, target_sum):
    """
    Check if there are any two distinct numbers within number_list that add up to target_sum.
    """
    seen = set()
    for number in number_list:
        if target_sum - number in seen:
            return True
        seen.add(number)
    return False

def run_tests_and_assert_complete(test_suite):
    results = unittest.TextTestRunner(verbosity=2).run(test_suite)
    assert results.wasSuccessful(), "Not all tests passed."

#<INSERT>

# 创建测试套件并添加测试用例
def create_test_suite():
    suite = unittest.TestSuite()
    suite.addTest(unittest.makeSuite(TestHasPairWithSum))
    return suite

# 运行测试并断言全部成功
test_suite = create_test_suite()
run_tests_and_assert_complete(test_suite)